Measurable Spaces – Problem (42/365)

The chapter on measurable spaces introduces a \sigma-algebra over the real numbers R = (-\infty, \infty). The Borel algebra, \mathcal{B}(R), is the smallest \sigma-algebra \sigma(\mathcal{A}) where \mathcal{A} is the algebra generated by finite disjoint sums of intervals of the form (a,b]. By the direct product of algebras we also get algebras over higher dimensions \mathcal{B}(R^n) = \mathcal{B}(R) \otimes \dots \otimes \mathcal{B}(R). We also get a legal \sigma-algebra for the infinite direct product \mathcal{B}(R^\infty) = \mathcal{B}(R) \otimes \mathcal{B}(R) \otimes \dots.

The book asks to show that certain sets are members of \mathcal{B}(R^\infty). Show that the following are Borel sets.

\displaystyle  \{ x \in R^\infty : \sup \inf x_n > a \} \\ \{ x \in R^\infty : \inf \sup x_n \le a \}

Take the first case. Note that \sup \inf x_n > a is not satisfied if for every i, \inf \{ x_k : k \ge i \} \le a. This can only happen if there are an infinite number of coordinates whose value is \le a. Let

\displaystyle  \text{let } I \subset \mathbb{N} \text{ finite subset of natural numbers} \\ S(I) = \cup_{i=1}^\infty \{ x \in \mathcal{B}(R^\infty) : x_i \le a \text{ if } i \in I \text{ and } x_i > a \text{ otherwise} \}

The set S(I) is a Borel set since we have constructed it as a countable union of Borel sets. Therefore, \{ x \in R^\infty : \sup \inf x_n > a \} = \cup_{I} S(I) (this is also a countable union) is a Borel set. A similar argument is made for the other.

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1 Response to Measurable Spaces – Problem (42/365)

  1. Pingback: Measurable Spaces – Problem (44/365) | Latent observations

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